Question: The lifespans of bears in a particular zoo are normally distributed. The average bear lives $46$ years; the standard deviation is $9.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living longer than $36.8$ years.
Explanation: $46$ $36.8$ $55.2$ $27.6$ $64.4$ $18.4$ $73.6$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $46$ years. We know the standard deviation is $9.2$ years, so one standard deviation below the mean is $36.8$ years and one standard deviation above the mean is $55.2$ years. Two standard deviations below the mean is $27.6$ years and two standard deviations above the mean is $64.4$ years. Three standard deviations below the mean is $18.4$ years and three standard deviations above the mean is $73.6$ years. We are interested in the probability of a bear living longer than $36.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the bears will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the bears will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $36.8$ years and the other half $({16\%})$ will live longer than $55.2$ years. The probability of a particular bear living longer than $36.8$ years is ${68\%} + {16\%}$, or $84\%$.